Given, Q(0, 2) is equidistant from P(5, -3) and R(x, 7), which means PQ = QR.
Find the distance of PQ and QR using distance formula,
\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
\(\begin{aligned} P Q=\sqrt{(5-0)^2+(-3-2)^2}=5 \sqrt{2} \end{aligned}\)
\(\begin{aligned} Q R=\sqrt{(0-x)^2+(2-7)^2}=\sqrt{x^2+25} \end{aligned}\)
Now, \(\begin{aligned} P Q=Q R \Rightarrow 5 \sqrt{2}=\sqrt{x^2+25} \end{aligned}\)
On squaring both sides of the above equation, we get
50 = x² + 25 \(\Rightarrow\) x² = 25  \(\Rightarrow\) x = \(\pm 5\)
Therefore, the values of x are 5 and -5.

Given points are A(2, - 4), P(3, 8) and Q(-10, y). 
According to the question, PA = QA
\(\sqrt{(2-3)^{2}+(-4-8)^{2}}=\sqrt{(2+10)^{2}+(-4-y)^{2}}\) 
\(\Rightarrow \ \sqrt{(-1)^{2}+(-12)^{2}}=\sqrt{(12)^{2}+(4+y)^{2}}\) 
\(\Rightarrow \ \sqrt{145}=\sqrt{160+y^{2}+8 y}\) 
On squaring both sides, we get
145=160+y²+8y
\(\Rightarrow \ y^{2}+8 y+160-145=0\) 
\(\Rightarrow \ y^{2}+8 y+15=0\) 
\(\begin{aligned} \Rightarrow &y=-5 \text { or } y=-3 \end{aligned}\) 
Now,\(P Q=\sqrt{(-10-3)^{2}+(y-8)^{2}}\) 
For \(y=-5, P Q=\sqrt{(-13)^{2}+(-5-8)^{2}}\) 
\(=\sqrt{338} \text { units }\) 
and for \(y=-3, P Q=\sqrt{(-13)^{2}+(-3-8)^{2}}\) 
\(=\sqrt{290} \text { units }\) 
Hence, the values of yare -5, -3 and distance is 
\(P Q=\sqrt{338} \text { units or } \sqrt{290} \text { units. }\).

Non-collinear

We know that, distance between two points P(x₁, y₁)
and \(Q\left(x_{2}, y_{2}\right) \text { is } \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\) 
Here, P(x_1,y_l) = P(x,y)andQ(x₂, y_z) = Q(O, 0)
\(\therefore\) Distance between point P (x, y) and origin is
\(\sqrt{(0-x)^{2}+(0-y)^{2}}=\sqrt{x^{2}+y^{2}}\) units

Let A=(a,1), B=(b,1) and C = (c,1), then prove AB = BC + CA

Here, A=(x₁,y₁)=(1, 3) and B=(x₂,y₂)=(a, 7)
Distance between A and B is AB = \(\sqrt { { \left( a-1 \right) }^{ 2 }+{ (7-3) }^{ 2 } } \)  [\(\because \) distance=\(\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }-{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } } \)]
\(\Rightarrow \)   5=\(\sqrt { { \left( a-1 \right) }^{ 2 }+{ (4) }^{ 2 } } \)      [\(\because \) distance, AB=5]
\(\Rightarrow \)   5=\(\sqrt { { a }^{ 2 }+1-2a+16 } \)
\(\Rightarrow \)  25=\({ a }^{ 2 }\)-2a+17        [squaring on both sides]
\(\Rightarrow \)  \({ a }^{ 2 }\)-2a-8=0    
\(\Rightarrow \)  \({ a }^{ 2 }\)-4a+2a-8=0
\(\Rightarrow \)  a(a-4)+2(a-4)=0
\(\Rightarrow \)  (a-4)(a+2) =0 \(\Rightarrow \)   a=4, -2
Hence, the possible values of a are 4 and -2.

(i) Considering A as origin (0, 0), AB as x-axis and AD as y-axis.
 Niharika runs in the 2nd line with green flag and distance covered (parallel to AD)
= \(\frac{1}{4} \times100 = 25 m\)
∴ Coordinates of green flag are (2, 25) and label it as P i.e., P(2, 25).
Similarly, Preet runs in the eighth line with red flag and distance covered (parallel to AD) 
= \(\frac {1}{5} \times100 = 20 m\)
∴ Coordinates of red flag are (8, 20) and label it as Q i.e., Q(8, 20).
Also, Rashmi has to post a blue flag at the mid- point of PQ, therefore, by using mid-point formula, we have \((\frac {2+8}{2} , \frac{25+20}{2})\)
i.e., \(\left( 5,\frac { 45 }{ 2 } \right) \)
Hence, the blue flag is in the fifth line  distance of \(\frac {45}{2}\) i.e., 22.5 m along the direction  parallel to AD.
(ii) Sports day is celebrated on 29th August, every year. 

Let the given points be L (b+a,c) , M(c+a,b) and N(c+b, a).
∴ By using distance formula, we get
LM = \(\sqrt {(c+a-b-a)^{2}+(b-c)^{2}}\)
=\(\sqrt{(c-b)^{2}+(c-b)^{2}(-1)^{2}}\)
=\(\sqrt{(c-b)^{2}+(c-b)^{2}}\)=(c-b)\(\sqrt{2}\)units
MN = \(\sqrt{(c+b-c-a)^{2}+(a-b)^{2}}\)
=\(\sqrt{(b-a)^{2}+(-1)^{2}+(b-a)^{2}}\)
=\(\sqrt{(b-a)^{2}+(b-a)^{2} }= (b-a)\sqrt{2}\)units
and LN = \(\sqrt {(c+b-b-a)^{2}+(a-c)^{2}}\)
=\(\sqrt{(c-a)^{2}+(-1)^{2}(c-a)^{2}}\)
=\(\sqrt{(c-a)^{2}+(c-a)^{2}}\)=(c-a)\(\sqrt{2}\)units
Now, LM+MN = (c-b)\(\sqrt{2}\) + (b-a) \(\sqrt{2}\)
= (c-b+b-a) \(\sqrt{2}\)
=(c-a)\(\sqrt{2}\)
=LN
Hence, the points (b+a, c), (c+a, b) and (c+b, a) are collinear.  

Let given points are P(1,5), Q(2,3) and R(-2,-11).
Now, PQ = \(\sqrt{(2-1)^{2}+(3-5)^{2}}\)
=\(\sqrt{(1)^{2}+(-2)^{2}}\)=\(\sqrt{1+4}\)
=\(\sqrt{5}\)units
QR = \(\sqrt{(-2-2)^{2}+(-11-3)^{2}}\)
=\(\sqrt{(-4)^{2}+(-14)^{2}}\) = \(\sqrt{16+196}\)
=\(\sqrt{212}\)=\(2\sqrt{53}\)units
PR = \(\sqrt{(-2-1)^{2}+(-11-5)^{2}}\)
=\(\sqrt{(-3)^{2}+(-16)^{2}}\) = \(\sqrt{9+256}\)
=\(\sqrt{265}\) = \(\sqrt {5} \times \sqrt {53}\) units 
Here, PQ+QR ≠ PR+QR ≠ PQ, PQ+PR ≠ QR
Hence, the points P, Q are not collinear.

Let P(x, 0) be the required point on x-axis.
∴ |PA|=|PB|
⇒ PA²=PB²
⇒ (x-5)²+(0-4)²=(x+2)²+(0-3)²
⇒ x²-10x+25+16=x²+4x+4+9
⇒ 14x=28
⇒ x=2
Hence, the required point is (2, 0).

Since (2, 0) and (-4, 0) both lie on x-axis, therefore, the required point on x-axis equidistant from (2, 0) and (-4, 0) is (-1, 0).

8 units

Points (a, b + c), (b, c + a) and (c,a + b) will be collinear if area of the triangle formed by these points is zero. 
Area = \(\frac{1}{2}\)[a{c+a-(a+b)}+b{a+b-(b+c)}+c{b+c-c{c+a}]
=\(\frac{1}{2}\)[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]
=\(\frac{1}{2}\)[ac-ab+ab-bc+bc-ac]
=\(\frac{1}{2}\)[0]=0
Thus, the given points are collinear.

scalene triangle

If \(\left[ { x }_{ 1 }\left( { y }_{ 2 }-{ y }_{ 3 } \right) +{ x }_{ 2 }\left( { y }_{ 3 }-{ y }_{ 1 } \right) +{ x }_{ 3 }\left( { y }_{ 1 }-{ y }_{ 2 } \right) \right] \)
\(\therefore \frac { -3 }{ 2 } \left( -2-4 \right) +6(4-3)+\left[ (-3)\left\{ 3-(-2) \right\} \right] =\frac { -3 }{ 2 } \times (-6)+6\times 1-3\times 5=9+6-15=0.\)
∴ P,Q and R are collinear.

\(AB=\sqrt { (4-{ 2) }^{ 2 }+(6-3)^{ 2 } } =\sqrt { 13 } ;\)
\(BC=\sqrt { (6-{ 4) }^{ 2 }+(9-6)^{ 2 } } =\sqrt { 13 } \)
\(AC=\sqrt { (6-{ 2) }^{ 2 }+(9-3)^{ 2 } } =\sqrt { 52 } =2\sqrt { 13 } \)
Here AB+BC=AC
∴ AB, BC and AC cannot be the side of any triangle.
( ∵ Sum of two sides of a triangle is always more than the third side)
∴  A, B and C are not vertices of any triangle. A, B and C are collinear.

\(AB=\sqrt { (14-{ 2) }^{ 2 }+(10+2)^{ 2 } } =12\sqrt { 2 } units\)
\(BC=\sqrt { (11-{ 14) }^{ 2 }+(13-10)^{ 2 } } =3\sqrt { 2 } units\)
\(CD=\sqrt { (-1-{ 11) }^{ 2 }+(1-13)^{ 2 } } =12\sqrt { 2 } units\)
\(AD=\sqrt { (-1-{ 2) }^{ 2 }+(1+2)^{ 2 } } =3\sqrt { 2 } units\)
⇒ AB=CD and BC = AD
∴ ABCD is a || gm
Now, \(AC=\sqrt { (11-2)^{ 2 }+({ 13+2) }^{ 2 } } =\sqrt { 306 } \)
⇒ AC²=306 units, AB² = 288 units, BC² =18 units
AB²+BC²=306 units
⇒ AC²=AB²+BC²⇒∠ABC = 90⁰
⇒ ABCD is a rectangle.

\(AB=\sqrt { (6-{ 3) }^{ 2 }+(0-5)^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
\(\\ BC=\sqrt { { (6-1) }^{ 2 }+{ (0+3) }^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
\(CD=\sqrt { (1+2)^{ 2 }+({ -3-2) }^{ 2 } } =\sqrt { 9+25 } =\sqrt { 34 } \)
\(\\ DA=\sqrt { (-2-3)^{ 2 }+({ 2-5) }^{ 2 } } =\sqrt { 25+9 } =\sqrt { 34 } \)
\(AC=\sqrt { (1-3)^{ 2 }+({ -3-5) }^{ 2 } } =\sqrt { 4+64 } =\sqrt { 68 } \)
\(BD=\sqrt { (6+2)^{ 2 }+({ 0-2) }^{ 2 } } =\sqrt { 64+4 } =\sqrt { 68 } \)
AB=BC=CD=DA=\(\sqrt{34}\)
Diagonal AC= diagonal BD=\(\sqrt {68}\)
Hence A, B, C and D are vertices of a square.

AB = \(\sqrt { \{ -5-{ (-3) }^{ 2 }\} +(-5-2)^{ 2 } } \)
\(=\sqrt { 4+49 } =\sqrt { 53 } \)
\(BC=\sqrt { \{ 2-{ (-5) }^{ 2 }\} +(-3-(-5)\} ^{ 2 } } =\sqrt { 53 } \)
\(\\ CD=\sqrt { { (4-2) }^{ 2 }+\{ { 4-(-3) }\} ^{ 2 } } =\sqrt { 53 } \)
\(AD=\sqrt { \{ 4-(-3){ \} }^{ 2 }+({ 4-2) }^{ 2 } } =\sqrt { 53 } \)
Now AB=CD, BC=AD
∴ A, B, C and D are vertices of a parallelogram.
Also AB=BC
∴ ABCD is a rhombus.

Let A(0,4), B(0,0) and C(3,0) are the vertices of △ABC
\(AB=\sqrt { ({ 0-0) }^{ 2 }+({ 0-4) }^{ 2 } } =4\)
\(BC=\sqrt { { (0-3) }^{ 2 }+({ 0-0) }^{ 2 } } =3\)
\(AC=\sqrt { { (3-0) }^{ 2 }+({ 0-4) }^{ 2 } } =5\)
Perimeter of △ABC = AB+BC+CA = 4+3+5 = 12 units.

Using distance formula
\(AB=\sqrt { ({ -3-2) }^{ 2 }+[8-({ -2)] }^{ 2 } } \)
\(=\sqrt { { 5 }^{ 2 }+10^{ 2 } } =\sqrt { 25+100 } \)
\(=\sqrt { 125 } =5\sqrt { 5 } units\)
\(BC=\sqrt { [{ -1-(-3)] }^{ 2 }+(4-8)^{ 2 } } \)
\(=\sqrt { { 2 }^{ 2 }+4^{ 2 } } =\sqrt { 4+16 } =\sqrt { 20 } \)
\(=2\sqrt { 5 } units\)
\(AC=\sqrt { ({ -1-2) }^{ 2 }+[4-(-2)]^{ 2 } } \)
\(=\sqrt { { 3 }^{ 2 }+6^{ 2 } } =\sqrt { 9+36 } =\sqrt { 45 } \)
\(=3\sqrt { 5 } units\)
\(\because \ 2\sqrt { 5 } +3\sqrt { 5 } =5\sqrt { 5 } \)
∴ Points are collinear. 

P(-6,7) be P(x₁,y₁) and Q(-1,-5) be (x₂,y₂).
Then, \(PQ=\sqrt { { ({ x }_{ 2 }-{ x }_{ 1 }) }^{ 2 }-{ ({ y }_{ 2 }-{ y }_{ 1 }) }^{ 2 }​​ } \)
\(=\sqrt { { \{ (-1)-(-6)\} }^{ 2 }-{ \{ (7)-(-5)\} }^{ 2 } } \)
\(=\sqrt { ({ -1+6) }^{ 2 }+({ 7+5) }^{ 2 } } =\sqrt { 25+144 } \)
\(\\ =\sqrt { 169 } =13\quad units\)

\(AB=\sqrt { ({ 1+2) }^{ 2 }+({ 2-3) }^{ 2 } } =\sqrt { 9+1 } =\sqrt { 10 } \)
\(BC=\sqrt { ({ 7-1) }^{ 2 }+({ 0-2) }^{ 2 } } =\sqrt { 36+4 } =\sqrt { 40 } =2\sqrt { 10 } \)
\(AC=\sqrt { ({ 7+2) }^{ 2 }+({ 0-3) }^{ 2 } } =\sqrt { 81+9 } =\sqrt { 90 } =3\sqrt { 10 } \)
Since  \(AB+BC=\sqrt { 10 } +2\sqrt { 10 } =(1+2)\sqrt { 10 } =3\sqrt { 10 } =AC\)
Hence, the points A, B and C are collinear.

Let A(7,6), B(-3,4)) be the given points and P(x,0) be the required point.
Since, P is equidistant from A and B, therefore,
AP = BP ⇒  AP²=BP²
⇒ (x-7)²+(0-6)²=(x+3)²+(0-4)² ⇒ x²+49-14x+36 = x²+9+6x+16
⇒ -14x-6x = 25-85    ⇒ -20x=-60
\(\Rightarrow x=\frac { -60 }{ -20 } =3\)
∴ Required point on axis is (3,0).

Let point on y-axis be (0,a)
Now distance of this point from (5,-2) equal to distance from point (-3,2)
i.e.,  \(\Rightarrow \sqrt { 5^{ 2 }+\left( -2-a \right) ^{ 2 } } =\sqrt { \left( 3 \right) ^{ 2 }+\left( a-2 \right) ^{ 2 } } \)
Squaring and simplifying, we get
25+4+a²+4a = 9+a²+4-4a ⇒ 8a=-16  ⇒ a=-2